Question: Find the number of functions $f : \mathbb{R} \to \mathbb{R}$ such that
\[f(x + y) f(x - y) = (f(x) + f(y))^2 - 4x^2 f(y)\]for all real numbers $x$ and $y.$
Setting $x = y = 0,$ we get
\[f(0)^2 = 4f(0)^2.\]Then $f(0)^2 = 0,$ so $f(0) = 0.$

Setting $x = y,$ we get
\[4f(x)^2 - 4x^2 f(x) = 0,\]so $f(x) (f(x) - x^2) = 0.$  This tells us that for each value of $x,$ either $f(x) = 0$ or $f(x) = x^2.$  (Note that it does not tell us that either $f(x) = 0$ for all $x,$ or $f(x) = x^2$ for all $x.$)

We can easily check that $f(x) = x^2$ satisfies the given functional equation.  Otherwise, there exists some nonzero real number $a$ such that $f(a) = 0.$  Setting $y = a,$ we get
\[f(x + a) f(x - a) = f(x)^2\]for all $x.$  Suppose there exists a real number $b$ such that $f(b) \neq 0.$  Then $f(b) = b^2.$  Substituting $x = b$ into the equation above, we get
\[f(b + a) f(b - a) = f(b)^2 = b^4.\]Since $f(b) = b^2 \neq 0,$ both $f(b + a)$ and $f(b - a)$ must be nonzero.  Therefore, $f(b + a) = (b + a)^2$ and $f(b - a) = (b - a)^2,$ and
\[(b + a)^2 (b - a)^2 = b^4.\]Expanding, we get $a^4 - 2a^2 b^2 + b^4 = b^4,$ so $a^4 - 2a^2 b^2 = 0$.  Then $a^2 (a^2 - 2b^2) = 0.$  Since $a$ is nonzero, $a^2 = 2b^2,$ which leads to $b = \pm \frac{a}{\sqrt{2}}.$

This tells us that if there exists some nonzero real number $a$ such that $f(a) = 0,$ then the only possible values of $x$ such that $f(x) \neq 0$ are $x = \pm \frac{a}{\sqrt{2}}.$  We must have that $f(x) = 0$ for all other values of $x.$  We can then choose a different value of $a'$ such that $f(a') = 0,$ which leads to $f(x) = 0$ for all $x$ other than $x = \pm \frac{a'}{\sqrt{2}}.$  This forces $f(x) = 0$ for all $x,$ which easily satisfies the given functional equation.

Therefore, there are only $\boxed{2}$ functions that work, namely $f(x) = 0$ and $f(x) = x^2.$